Let $ABC$ a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$ and $E$ the foot of the bisector of the angle $A$. If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of the triangle $ABC$ and $\{F\} = DI \cap AM$, prove that $MI$ passes through the midpoint of $[EF]$.

Let $ABC$ be a triangle and let $O$ be the centre of its circumscribed circle. Points $X, Y$ which are neither of the points $A, B$ or $C$, lie on the circumscribed circle and are so that the angles $XOY$ and $BAC$ are equal (with the same orientation). Show that the orthocentre of the triangle that is formed by the lines $BY, CX$ and $XY$ is a fixed point.

In right-angle triangle $ABC$, $\angle C=90$°, Point $D$ is the midpoint of side $AB$. Points $M$ and $C$ lie on the same side of $AB$ such that $MB\bot AB$, line $MD$ intersects side $AC$ at $N$, line $MC$ intersects side $AB$ at $E$. Show that $\angle DBN=\angle BCE$.

Let $ ABC$ be an acute triangle. Take points $ P$ and $ Q$ inside $ AB$ and $ AC$, respectively, such that $ BPQC$ is cyclic. The circumcircle of $ ABQ$ intersects $ BC$ again in $ S$ and the circumcircle of $ APC$ intersects $ BC$ again in $ R$, $ PR$ and $ QS$ intersect again in $ L$. Prove that the intersection of $ AL$ and $ BC$ does not depend on the selection of $ P$ and $ Q$.

Let $ABC$ be a triangle, and $I$ its incenter. Consider a circle which lies inside the circumcircle of triangle $ABC$ and touches it, and which also touches the sides $CA$ and $BC$ of triangle $ABC$ at the points $D$ and $E$, respectively. Show that the point $I$ is the midpoint of the segment $DE$.

Points $A, B,$ and $C$ are on a circle such that $\triangle ABC$ is an acute triangle. $X, Y ,$ and $Z$ are on the circle such that $AX$ is perpendicular to $BC$ at $D$, $BY$ is perpendicular to $AC$ at $E$, and $CZ$ is perpendicular to $AB$ at $F$. Find the value of \[ \frac{AX}{AD}+\frac{BY}{BE}+\frac{CZ}{CF}, \] and prove that this value is the same for all possible $A, B, C$ on the circle such that $\triangle ABC$ is acute. [asy] pathpen = linewidth(0.7); pair B = (0,0), C = (10,0), A = (2.5,8); path cir = circumcircle(A,B,C); pair D = foot(A,B,C), E = foot(B,A,C), F = foot(C,A,B), X = IP(D--2*D-A,cir), Y = IP(E--2*E-B,cir), Z = IP(F--2*F-C,cir); D(MP("A",A,N)--MP("B",B,SW)--MP("C",C,SE)--cycle); D(cir); D(A--MP("X",X)); D(B--MP("Y",Y,NE)); D(C--MP("Z",Z,NW)); D(rightanglemark(B,F,C,12)); D(rightanglemark(A,D,B,12)); D(rightanglemark(B,E,C,12));[/asy]

In triangle $ABC$ points $P$ and $Q$ lies on the external bisector of $\angle A$ such that $B$ and $P$ lies on the same side of $AC$. Perpendicular from $P$ to $AB$ and $Q$ to $AC$ intersect at $X$. Points $P'$ and $Q'$ lies on $PB$ and $QC$ such that $PX=P'X$ and $QX=Q'X$. Point $T$ is the midpoint of arc $BC$ (does not contain $A$) of the circumcircle of $ABC$. Prove that $P',Q'$ and $T$ are collinear if and only if $\angle PBA+\angle QCA=90^{\circ}$.

It is known that $ \angle BAC$ is the smallest angle in the triangle $ ABC$. The points $ B$ and $ C$ divide the circumcircle of the triangle into two arcs. Let $ U$ be an interior point of the arc between $ B$ and $ C$ which does not contain $ A$. The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AU$ at $ V$ and $ W$, respectively. The lines $ BV$ and $ CW$ meet at $ T$. Show that $ AU \equal{} TB \plus{} TC$. [i]Alternative formulation:[/i] Four different points $ A,B,C,D$ are chosen on a circle $ \Gamma$ such that the triangle $ BCD$ is not right-angled. Prove that: (a) The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AD$ at certain points $ W$ and $ V,$ respectively, and that the lines $ CV$ and $ BW$ meet at a certain point $ T.$ (b) The length of one of the line segments $ AD, BT,$ and $ CT$ is the sum of the lengths of the other two.

ABC is an isosceles triangle with AB=AC. Point D lies on side BC. Point F is inside $\triangle$ABC and lies on the circumcircle of triangle ADC. The circumcircle of triangle BDF intersects side AB at point E. Prove that $CD\cdot EF+DF\cdot AE=BD\cdot AF$.

Let $ ABC$ be a triangle. A circle $ P$ is internally tangent to the circumcircle of triangle $ ABC$ at $ A$ and tangent to $ BC$ at $ D$. Let $ AD$ meets the circumcircle of $ ABC$ agin at $ Q$. Let $ O$ be the circumcenter of triangle $ ABC$. If the line $ AO$ bisects $ \angle DAC$, prove that the circle centered at $ Q$ passing through $ B$, circle $ P$, and the perpendicular line of $ AD$ from $ B$, are all concurrent.

$ABC$ is any triangle. Tangent at $C$ to circumcircle ($O$) of $ABC$ meets $AB$ at $M$. Line perpendicular to $OM$ at $M$ intersects $BC$ at $P$ and $AC$ at $Q$. P.T. $MP=MQ$.

Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABO$ and $ACO$, respectively. The circumcircle of $\triangle AO_1O_2$ intersects segment $BC$ at two distinct points $P$ and $Q$, with the four points $B, P, Q, C$ appearing in this order along $BC$. Let $O_3$ be the circumcenter of $\triangle OPQ$. Prove that points $A, O, O_3$ are collinear.

Let $ABC$ be a triangle, and $A_1, B_1, C_1$ be points on the sides $BC, CA, AB$, respectively, such that $AA_1, BB_1, CC_1$ are the internal angle bisectors of $\triangle ABC$. The circumcircle $k' = (A_1B_1C_1)$ touches the side $BC$ at $A_1$. Let $B_2$ and $C_2$, respectively, be the second intersection points of $k'$ with lines $AC$ and $AB$. Prove that $|AB| = |AC|$ or $|AC_1| = |AB_2|$.

Let $ABCD$ be a parallelogram and $P$ be a point on diagonal $BD$ such that $\angle PCB=\angle ACD$. Circumcircle of triangle $ABD$ intersects line $AC$ at points $A$ and $E$. Prove that \[\angle AED=\angle PEB.\]

Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right. [i]L. Kuptsov[/i]

In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .

Points $P,Q,R,S$ lie on a circle and $\angle PSR$ is right. $H,K$ are the projections of $Q$ on lines $PR,PS$. Prove that $HK$ bisects segment $ QS$.

Let $ABC$ be a triangle. Let $H$ be the foot of the altitude from $C$ on $AB$. Suppose that $AH = 3HB$. Suppose in addition we are given that (a) $M$ is the midpoint of $AB$; (b) $N$ is the midpoint of $AC$; (c) $P$ is a point on the opposite side of $B$ with respect to the line $AC$ such that $NP = NC$ and $PC = CB$. Prove that $\angle APM = \angle PBA$.

Let $ P$, $ Q$, and $ R$ be the points on sides $ BC$, $ CA$, and $ AB$ of an acute triangle $ ABC$ such that triangle $ PQR$ is equilateral and has minimal area among all such equilateral triangles. Prove that the perpendiculars from $ A$ to line $ QR$, from $ B$ to line $ RP$, and from $ C$ to line $ PQ$ are concurrent.

Let $I$ be the incentre of a triangle $ABC$ and $\Gamma_a$ be the excircle opposite $A$ touching $BC$ at $D$. If $ID$ meets $\Gamma_a$ again at $S$, prove that $DS$ bisects $\angle BSC$.

In $\triangle ABC$, $AB>AC$. In the circumcircle $(O)$ of $\triangle ABC$, $M$ is the midpoint of arc $BAC$. The incircle $(I)$ of $\triangle ABC$ touches $BC$ at $D$, the line through $D$ parallel to $AI$ intersects $(I)$ again at $P$. Prove that $AP$ and $IM$ intersect at a point on $(O)$.

Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BH_b, CH_c$ intersect at orthocenter $H$ ($H_b \in AC$, $H_c \in AB$). $H_bH_c$ meets $BC$ at $P$. Let $N$ be the midpoint of $AH, L$ be the orthogonal projection of $O$ on the symmedian with respect to angle $A$ of triangle $ABC$. Prove that $\angle NLP = 90^o$.

In equilateral $\triangle ABC$, point $D$ lies on $\overline{BC}$ such that the radius of the circumcircle $\Gamma_1$ of $\triangle ACD$ is $7$ and the radius of the incircle $\Gamma_2$ of $\triangle{ABD}$ is $2$. Suppose that $\Gamma_1$ and $\Gamma_2$ intersect at points $X$ and $Y$. Find $XY$.

In triangle $ABC$ it holds $|BC|= \frac{1}{2}(|AB|+|AC|)$. Let $M$ and $N$ be midpoints of $AB$ and $AC$, and let $I$ be the incenter of $ABC$. Prove that $A, M, I, N$ are concyclic.

The bisector of $\angle{BAD}$ in the parallellogram $ABCD$ intersects the lines $BC$ and $CD$ at the points $K$ and $L$ respectively. Prove that the centre of the circle passing through the points $C,\ K$ and $L$ lies on the circle passing through the points $B,\ C$ and $D$.